Bogachev in his book on measure theory provided the criteria of integrability of a function $f$ with respect to a finite nonnegative measure $\mu$ viz. $\sum_{n=1}^\infty\mu(x: |f(x) |\geq n)$ converges.

Then he stated two examples, one of which is (emphasis mine):

A function $f$ measurable with respect to a bounded measure $\mu$ is integrable in every degree $p\in(0, \infty) $ precisely when the function $\mu(x: |f(x) |\geq t)$ decreases faster than any power of $t$ as $t\to + \infty.$

Author in another of his book however stated

... decreases faster than every negative power of $t$ ...

I couldn't grasp what he meant by "decreases faster". What did he try to imply by this? Does "negative" change the discussion in any way?

  • $\begingroup$ I think that he meant that if u take $\frac{1}{t^n}$, and divide it by μ(x:|f(x)|≥t). and taking $t \rightarrow \infty$. you will get zero for the limit for every $n \in \mathbb{N}$. Using O notation : $\mu = O(t^n) \text{ for every } n \in \mathbb{N}$. $\endgroup$ Sep 19, 2023 at 16:15

1 Answer 1


Decreases faster than any power of $t$ would typically mean that $$\lim_{t\to\infty} t^a \mu(\{x \, : \, \vert f(x) \vert \ge t\}) = 0, \,\,\,\,\, \text{ for all } \,\, a >0. \tag 1$$ To prove the statement, you can then use the "layer cake" representation of the $p$-norm: $$\|f\|_p^p = p\int^\infty_0 t^{p-1} \mu(\{x \, : \, \vert f(x) \vert \ge t\}) dt. \tag 2$$ If statement $(1)$ is true, then apply the statement for $a = p+1$, we find that for sufficiently large $t$, $$t^{p-1}\mu(\{x \, : \, \vert f(x) \vert \ge t\}) \le \frac 1 {t^2}$$ so plugging this bound into $(2)$ proves that $f$ is $p$-integrable.

  • $\begingroup$ Thanks for the answer, which gave me some insight of what's happening. But few queries: after checking another book by the same author, I noticed this time he mentioned "decreases faster than every negative power of $t$ ... " Does it make a difference? // plugging the bound in $(2)$ gives $\int f^p\leq\int_0^\infty 1/t^2~\mathrm dt $ - doesn't it diverge? // how does the theorem that integrability implies the convergence of $\sum_{n=1}^\infty\mu(x: |f(x) |\geq t)$ come to play? // Appreciate your feedback. $\endgroup$ Sep 19, 2023 at 19:18
  • 1
    $\begingroup$ @User1865345 Regarding the "decreases faster than any negative power of $t$": no that doesn't make a difference. The author means $\mu(\{\vert f \vert \ge t\}) = o(t^{-a})$ for all $a > 0$ which is equivalent to my statement above. Regarding using the bound in $(2)$: because the measure is finite, you can use the bound $\mu(\{\vert f \vert \ge t\}) \le \mu(X)$ for $0\le t \le 1$, and use the bound provided for $t>1$, and then there is no issue of divergence. For your last statement, the summand doesn't depend on $n$, so I think there is a typo. If you clarify, I would be happy to try to help $\endgroup$
    – User8128
    Sep 21, 2023 at 15:45
  • $\begingroup$ Thanks for the comment. My bad; the last query involved $n$ in place of $t.$ I would edit my post too. Let me read your whole comment then and let you know. $\endgroup$ Sep 21, 2023 at 15:50
  • $\begingroup$ I think your comment helped clearing the confusions regarding the first two queries (thanks!). However, could you tell me about my third query? I am failing to see how the author in this example used the criteria that $\sum_{n=1}^\infty\mu(x: |f(x) |\geq n)<\infty$ would ensure the integrability of $|f|.$ After all, it was one of the examples showing the use of the theorem. $\endgroup$ Sep 22, 2023 at 3:31

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